When a body hanging from a spring balance is immersed in water it floats just below the surface of water and shows 100g loss in its weight (I) what is the weight of the body in water? (Ii) what is the weight of the body in air? (Iii) what is the up thrust on the body? (Iv) what is the volume of the body?
50 gm,0gm,100gm,100gm
(I) 50 gm (ii) 0gm (iii) 100 gm (iv) 100gm
which law of chemical combination is illustrated by the fact that 168g Fe reacts with 72g H2o to form 232g Fe3O4 and8g H2
To find the answers to these questions, we will use the principles of buoyancy and Archimedes' principle.
(I) The weight of the body in water can be found by subtracting the loss in weight from its original weight in air. The loss in weight is given as 100g, so if the body weighed W in air, then the weight in water would be W - 100g.
(II) The weight of the body in air is the same as its original weight. So, the weight of the body in air is W.
(III) The upthrust on the body is equal to the weight of the water displaced by the body. According to Archimedes' principle, when an object is immersed in a fluid, it experiences an upward force equal to the weight of the fluid displaced. Therefore, the upthrust is also equal to the loss in weight, which is 100g.
(IV) The volume of the body can be calculated using the density of water and the loss in weight. The formula for calculating the volume of an object is: Volume = (Loss in weight) / (Density of water). Since the loss in weight is given as 100g, and the density of water is approximately 1g/cm^3, the volume of the body would be 100 cm^3.
So, to summarize:
(I) Weight of the body in water = W - 100g.
(II) Weight of the body in air = W.
(III) Upthrust on the body = 100g.
(IV) Volume of the body = 100 cm^3.